Try your hand at this sample problem focusing on a specific probability situation.

Sample Problem:

Each person in Room A is a student, and 1/6 of the students in Room A are seniors. Each person in Room B is a student, and 5/7 of the students in Room B are seniors. If 1 student is chosen at random from Room A and 1 student is chosen at random from Room B, what is the probability that exactly 1 of the students chosen is a senior?

(A) 5/42
(B) 37/84
(C) 9/14
(D) 16/21
(E) 37/42

Solution:

In this problem we are asked to determine the probability of choosing exactly 1 senior. To be more specific, this means that we will need to select 1 senior and 1 non-senior.

There are two ways this can be done. We can select a senior from Room A and a non-senior from Room B or we can select a non-senior from Room A and a senior from Room B. In probability, ‘and’ means multiply while ‘or’ means add. In this case, we see 2 ‘ands’ indicating that we will multiply probabilities together and one ‘or’ telling us that we will add the resulting products.

Our first option is to get a senior from Room A, which has a probability of 1/6, and a non-senior from Room B, which has a probability of 2/7 (we found this number by taking the probability of selecting a senior in Room B and subtracting it from 1.) Because this is an ‘and’ situation, we multiply 1/6 * 2/7, which equals 2/42.

For our other option we must find the probability of selecting a non-senior from Room A, which has a probability of 5/6, and senior from Room B, which has a probability of 5/7. Again, we have an ‘and’ statement, so we multiply. 5/6 * 5/7 = 25/42.

Lastly, we must address the ‘or’ in our original scenario by adding the two probabilities we determined together. This gives us 2/42 + 25/42 = 27/42. This simplifies to 9/14, which is choice (C).